Question: The grades on a physics midterm at Covington are normally distributed with $\mu = 84$ and $\sigma = 2.0$. Jessica earned a $79$ on the exam. Find the z-score for Jessica's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Jessica's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{79 - {84}}{{2.0}}} $ ${ z \approx -2.50}$ The z-score is $-2.50$. In other words, Jessica's score was $2.50$ standard deviations below the mean.